I would use a FEM code (http://en.wikipedia.org/wiki/Finite_element_method_in_structural_mechanics).
For a simple rule of thumb method, I have to know the geometry of the rig. For a beach cat, you could assume to have the weight of crew, helmsman and half boat weight on the shroud, apply a margin of safety of 3...4, then you have a rough idea of the loading. E.g. 2-up beach cat of 100kg weight: (2 x 80 + 50) x 3 = 630 kg. This requires at least a 3mm wire. Keep in mind that a margin of 3 is low, and requires a regular check. Lazy guys could go with a 4mm wire and less regular checking.

Cheers,

Klaus

The answer is "8"

It's pretty straight forward to approximate the upwind load on the shroud if you only look at the system in two dimensions. When you take into consideration that the shroud actually mounts to the hull slightly aft of the mast base, the calculation becomes trickier because now you are dealing with a system in three dimensions.

But, to get a fairly good approximation, the basic trick is to realize that for the boat at a constant angle of heel (and thus zero rotational acceleration) the righting moment is equal to the heeling moment. So determine the maximum righting moment by multiplying the center of mass of the boat & crew by the horizontal distance of the center of mass from the leeward hull (fulcrum). Then, take this number and divide it by the vertical distance from the mast base to the mast tang. This will give you the horizontal component of the force required to support the mast. Using trigonmetry, you can then determine the force along the shroud based on the horizontal force and the geometry of the shroud connection points. The force along the shroud will be equal to the horizontal component of shroud load divided by the cosine of the angle between the deck and the shroud.

This gives you a rough (but probably fairly accurate) load for the shroud while sailing upwind with no one on the trapeze. Downwind would be a little different because in that case, there is forward load on the mast and also, the bouyancy of the bows adds to the righting moment. This also doesn't take into account "spikes" in the shroud loading that results from sailing over chop/rough conditions.

sm

Not that hard at all. Why do you need it ?

But for the downwind case, I would say that the wind forces are what come to play, and the angle of the shrouds becomes a big amplifier of the force in the forward direction. Similar story for the forestay, it´s the mostly the will wind that determine the stay tension. Which is the worst case?

The safety factor for steel wires can be smaller than 3 to 4. I guess the high figures avoid the need to calculate dynamic loads, which can be very important.

I mainly ask for a rough educated guess based on some science.... My boat came with 9mm dynex dux synthetic stays which have a breaking strength of 27k lbs. I was mainly curious about what the "safety factor" is to account for some possible unforseen abrasion on the synthetic stays.